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exponent challenge

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exponent challenge

by chrishoward76 » Wed Feb 20, 2008 5:59 pm
Hey, just getting my math feet back under me, I am not so bad at exponents, but this question is bugging me. Not sure how I could quickly solve it.
can some one show me a fast way?

(5^21)(4^11)=2x10^n
n=21
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Re: exponent challenge

by Stuart@KaplanGMAT » Wed Feb 20, 2008 7:08 pm
chrishoward76 wrote:Hey, just getting my math feet back under me, I am not so bad at exponents, but this question is bugging me. Not sure how I could quickly solve it.
can some one show me a fast way?

(5^21)(4^11)=2x10^n
n=21
We see "5" and "4" on the left side and "10" on the right side.. somehow we want to make the bases the same.

So, let's start by turning the "4" into "2^2":

(5^21)((2^2)^11) = 2*10^n

(5^21)(2^22) = 2*10^n

Now, we want to be able to simplify the left side, but we can only do so if the exponents are equal, so:

(5^21)(2^21)(2^1) = 2*10^n

if we divide both sides by 2, we get:

(5^21)(2^21) = 10^n

and since the exponents are equal on the left side, we can say:

(5*2)^21 = 10^21

and finally

10^21 = 10^n

so 21 = n
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by chrishoward76 » Wed Feb 20, 2008 7:15 pm
Stuart- thank you very much for that quick response- totally makes sense!
Cheers
Chris
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by khurram » Sat May 24, 2008 10:09 pm
great explanation Stuart.

Thanks
Khurram
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